classSolution{public:vector<int>largestDivisibleSubset(vector<int>& nums){intl = nums.size();if(l==0||l==1)returnnums;std::vector<int>dp(l,0),parent(l,-1);sort(nums.begin(), nums.end());intbegin =0;intmaxdp =0;for(inti = l-1; i >=0; --i) {for(intj = i; j < l...
368. Largest Divisible Subset FindBorderBarSize Given a set ofdistinctpositive integersnums, return the largest subsetanswersuch that every pair(answer[i], answer[j])of elements in this subset satisfies: answer[i] % answer[j] == 0, or answer[j] % answer[i] == 0 If there are multiple...
1publicclassSolution {2publicstaticList<Integer> largestDivisibleSubset(int[] nums) {34if(nums.length==0){5returnnewArrayList<Integer>();6}7if(nums.length==1){8List<Integer> array =newArrayList<Integer>();9array.add(nums[0]);10returnarray;11}12Arrays.sort(nums);13intlen =nums.length;14...
Largest Divisible Subset Desicription Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0. If there are multiple solutions, return any subset is fine. Example 1: 代码语言:...
AI代码解释 publicList<Integer>largestDivisibleSubset(int[]nums){int[]count=newint[nums.length];int[]pre=newint[nums.length];Arrays.sort(nums);int maxIndex=-1;int max=0;for(int i=0;i<nums.length;i++){count[i]=1;pre[i]=-1;for(int j=i-1;j>=0;j--){if(nums[i]%nums[j]=...
mx=dp[i].first; mx_idx=i; } } } }for(inti =0; i < mx; ++i) { res.push_back(nums[mx_idx]); mx_idx=dp[mx_idx].second; }returnres; } }; 最大可整除的子集合[LeetCode] Largest Divisible Subset,如需转载请自行联系原博主。
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0. If there are multiple solutions, return any subset is fine.
368 Largest Divisible Subset // #368 最大整除子集 描述:给定一个集合,找出最大的子集,使得子集中任意两数都有整除关系。 //#368Description: Largest Divisible Subset | LeetCode OJ 解法1:DP. // Solution 1: DP. 代码1 //Code 1 369 Plus One Linked List ...
publicList<Integer>largestDivisibleSubset(int[] nums) { intlen = nums.length; Arrays.sort(nums); // 第 1 步:动态规划找出最大子集的个数、最大子集中的最大整数 int[] dp =newint[len]; Arrays.fill(dp,1); intmaxSize =1; intmaxVal = dp[0]; ...
1importjava.util.*;23classSolution {4publicstaticList<Integer> largestDivisibleSubset(int[] nums) {5List<Integer> result=newArrayList<Integer>();6List<Integer> numsList=newArrayList<Integer>();7for(inti=0;i<nums.length;i++){8numsList.add(nums[i]);9}10if(nums.length==0)returnresult;11...